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Algebra

Scorching goIITian

Joined:	27 Sep 2007
Post:	235

28 Dec 2007 07:30:39 IST

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379

sequence and progression!!!!!!!!!!!!!!!!!!!!!

None

Q 1. an ap and an hp have the same first term the same last term and the same no. of terms......prove that the product of the rth term from the beginning in one series and the rth term from the end in the other is independent of r

Q 2. given a^x=b^y=c^z=d^u and a,b,c,d are in gp,show that x,y,z,u are in hp.

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Comments (4)

Bhaskar Tetarbe

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Joined: 15 Jul 2007

Posts: 245

28 Dec 2007 09:44:09 IST

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Hey, the first question is a bit long one so I ve solved Q2.
given a,b,c,d in G.P
Let us say that r is the common ratio.
then , a^x=(ar)^y=(ar²)^z=(ar³)^u=k (say)
Then, a = k^{1/x                    -------------------------(1)}        ar = k^1/y-------------------(2)
       ar² = k^1/z              --------------------(3)
       ar³ = k^{1/u                  ---------------------------(4)}Divide, (2) by (1),
r         =   k^{1/y - 1/x}
Divide, (3) by (2),
r         =   k^{1/z - 1/y}Divide, (4) by (3),
r         =   k^{1/u - 1/z}Therefore, 1/y - 1/x = 1/z - 1/y = 1/u - 1/z
Hence, x,y,z,u are in H.P.

hitesh sharma

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28 Dec 2007 10:35:41 IST

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thanx yaar it was so easy ....and seriously i am a dumbo....waise thanx so much...could u tell me just the technique of 1st question i mean hint

hitesh sharma

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Joined: 27 Sep 2007

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28 Dec 2007 10:45:09 IST

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i know to solve the first question we just have to convert r in some other relation......par pata nahi aaj dimaak bilkul nahi chal raha

dont u know me

Blazing goIITian

Joined: 5 Dec 2007

Posts: 704

29 Dec 2007 23:34:35 IST

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good btetarbe brilliant job

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